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frutty [35]
3 years ago
12

3x-4y=11 3x+2y=2please help, I need to find what the answer is for x and y​

Mathematics
1 answer:
Zielflug [23.3K]3 years ago
3 0

Answer:

I have attached a picture of me solving the equation.

Now you must be wondering how I got the answer, well first simplify the expression in y = mx + b. Then graph it on demos. Do the same thing for the other expression. Pick any point that lies on the line. That is your solution to the equation.

The green line is the expression of 3x - 4y = 11. I simplify the equation in y = mx + b giving me y = 3/4x - 11/4.

The orange line is the expression of 3x + 2y = 2. I simplify the equation in y = mx + b giving me y = -3/2x + 1.

Hope this helps, thank you !!

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(60 divided by (3+7) x2 ) x 5 +7 A. 60 B 22 C 30 D 67
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Answer:

the answer is D

Step-by-step explanation:

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Serious explination on this one with the answer guys? First to answer gets brainliest!
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Answer:

the answer is -5x + -8x-5/2x^2-3

Step-by-step explanation:

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If f(x) = 2x2^2 + 1 and g(x) = x^2 - 7, find (f+g) (x)
Sveta_85 [38]

(f+g)(x) = f(x) + g(x)

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5 0
3 years ago
If a random sample of size nequals=6 is taken from a​ population, what is required in order to say that the sampling distributio
goldenfox [79]

Answer:

\bar X \sim N(\mu , \frac{\sigma}{\sqrt{n}})

In order to satisfy this distribution we need that each observation on this case comes from a normal distribution, because since the sample size is not large enough we can't apply the central limit theorem.

Step-by-step explanation:

For this case we have that the sample size is n =6

The sample man is defined as :

\bar X = \frac{\sum_{i=1}^n X_i}{n}

And we want a normal distribution for the sample mean

\bar X \sim N(\mu , \frac{\sigma}{\sqrt{n}})

In order to satisfy this distribution we need that each observation on this case comes from a normal distribution, because since the sample size is not large enough we can't apply the central limit theorem.

So for this case we need to satisfy the following condition:

X_i \sim N(\mu , \sigma), i=1,2,...,n

Because if we find the parameters we got:

E(\bar X) =\frac{1}{n} \sum_{i=1}^n E(X_i) = \frac{n\mu}{n}=\mu

Var(\bar X)= \frac{1}{n^2} \sum_{i=1}^n Var(X_i) = \frac{n\sigma^2}{n^2}= \frac{\sigma^2}{n}

And the deviation would be:

Sd (\bar X) = \frac{\sigma}{\sqrt{n}}

And we satisfy the condition:

\bar X \sim N(\mu , \frac{\sigma}{\sqrt{n}})

3 0
3 years ago
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